Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.Example:

Given the sorted linked list: [-10,-3,0,5,9],One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:      0     / \   -3   9   /   / -10  5

难度：medium

题目：给定一个单链表其元素为升序排列，将其转换成高度平衡的二叉搜索树

思路：中序遍历

Runtime: 1 ms, faster than 99.17% of Java online submissions for Convert Sorted List to Binary Search Tree.

Memory Usage: 41 MB, less than 9.56% of Java online submissions for Convert Sorted List to Binary Search Tree.

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } *//** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */class Solution {    public TreeNode sortedListToBST(ListNode head) {        if (null == head) {            return null;        }        ListNode ptr = head;        int count = 0;        for (; ptr != null; ptr = ptr.next, count++);        ListNode[] headList = {head};                return sortedListToBST(headList, 0, count - 1);    }        public TreeNode sortedListToBST(ListNode[] head, int start, int end) {        if (start > end) {            return null;        }        int mid = start + (end - start) / 2;        TreeNode left = sortedListToBST(head, start, mid - 1);        TreeNode root = new TreeNode(head[0].val);        root.left = left;        head[0] = head[0].next;                root.right = sortedListToBST(head, mid + 1, end);        return root;    }}